# Lines and planes in space

The equation for a straight line is only defined for two dimensions. It does not however stop them from existing in higher dimensions, where we define them in parametric or vector form. ## Intro

The fact that the concept of a straight line appears as one of the five opening axioms of Euclid's "The Element" suggests both its simplicity and power.

His work, which introduced rigorous mathematics to lines and planes, is often called the most influential textbook of all time.

It's possible that Euclid never existed as an individual, but was a pseudonym for a team of mathematicians.

For being the author of a book that is so highly praised, we know very little of Euclid of Alexandria. The lack of biographical information about him is unusual for Greek scientists of the third century B.C. when he was active.

So unusual in fact, that some historians have suggested the name is a pseudonym used by a team of mathematicians. There appear to be even less evidence to support this claim, however, than there is to justify Euclid as the one and only founding father of geometry.

## Concept

Points have no shape or size, but merely define a location in space. Yet, this property is enough to define many interesting structures, such as lines and planes.

These simple objects are some of the fundamental cornerstones of all geometry, which we use to visualize and interpret the world around us.

A line is a connected set of points in 1D that extends infinitely in both directions. Hence, a line can be fully described using only two points on it, since all others must follow the same straight path. Similarly, to define a plane, which is a 2D surface of points extending infinitely in every direction, three points are sufficient as long as they do not lie on the same line. ## Math

As an alternative to consider multiple points to describe a line or a plane, we can also use what is called a parameterization.

Parameterization uses a single point on the object, along with direction vectors indicating the directions in which the objects spread through the vector space.

The term parameterization comes from the fact that its equations use parameters to scale direction vectors. These are added to produce points on the object.

As the parameters go through all possible real numbers, the parameterization will go through all points belonging to the object.

Parameterization for a line takes on the form:

While for a plane, the form of a parameterization is:

Where p is a point on the respective objects, and are corresponding direction vectors, while s and t are parameters.

## Distance in space

A natural need is to be able to calculate distances in the space. It can be the distance between two points, between a point and a line or between a line and a plane. There are distance formulas for the different cases, and they are presented here.

Distance is always measured between two points. The distance formula is equivalent to creating a vector between two points and calculating its length.

It is tempting to memorize these, but it is not recommended because there is a strong connection between students who do not pass the exam and those who memorize these formulas instead of learning the craft.

### Distance between two points By a distance , we always mean the shortest distance between two points, which is equivalent to calculating the length of a vector that has been created from these two points. The following function applies for :

which can be recognized from the length of a vector. Since each vector is a relative difference in the different terms, that is, an arrow between two points, the formula for the distance between the points can be derived from the definition of the length of a vector.
The following applies to the distance formula :

Let and be points in the space . Then:

• if and only if

### Distance between a point and a line We will cover two methods to calculate the distance between a point and a line .

#### Method 1: Cross product in Let:

Take a point located on the line . With a little creativity, you can see that and span a parallelogram with the height . The area is the base multiplied by the height, which is also equal to the length of the cross product . Therefore, we can divide the latter by the length of the base (the vector ) to get the height .

The above formula only works for because the cross product is only defined then.

#### Method 2: Projection Let:

The distance can be found by projection, a relationship which is illustrated by a simple vector summation:

Therefore, it applies that:

#### Method 3: Dot product Consider the shortest distance between the point and the point on the line whose vector forms a right angle with the line. If both and are known points, the calculation of the length is simple, namely:

Remember that , which means that the vector is orthogonal to the line . We find the unknown point by setting the corresponding requirement for , i.e. the equation:

Because the two vectors form a right angle, the dot product becomes 0. The point is unknown with several coordinates, which results in several unknowns (one per coordinate) while we only have one equation. However, the point can be expressed using the parametric form for for some unknown parameter value :

where and are known. We make the following substitution:

The equation is solved for the as yet unknown value, namely , since the vector forms a right angle with . When solved, insert in the parametric form and get the desired point . Then we have two known points, and , and the distance between them is the length of their vector:

#### Method 4: Create a plane With a little creativity, we can construct the plane that contains the point and has a normal vector parallel to the direction vector for . From there, the intersection point between and can be determined, which will also be the closest point on the line to . The procedure is as follows:

1. Create the equation:

for the plane with normal vector = direction vector for = .

2. The constant is obtained by inserting the point into the plane's equation.

3. Insert the parametric form for into the plane . Solve for the parameter (we have one equation and one variable).

4. Let be the parameter that represents the intersection point . Insert in the parametric form for and get .

5. Calculate the distance:

### Distance between a point and a plane Let be a plane and be a point in the space as follows:

Create the line that passes through the point and hits the plane at a right angle. The direction vector therefore becomes the plane's normal vector (which we read from the plane's equation). We then get:

We are looking for the point of intersection between the line and the plane . The parametric form for expresses all points along the line, and we can express each such point in the following way:

Our design of guarantees an intersection point precisely because the direction vector is the plane's normal vector. Therefore, we put the above expression into the plane's equation:

The above equation is easily solved because only is unknown. Let's note the solution as :

which also returns the desired intersection point when it is inserted into the parametric form for . The point therefore has the following coordinates:

Now both and are known, and therefore the distance is easily calculated by creating a vector and calculating the length:

Which leads to the formula for the distance between a point and a plane. Again - it is recommended to remember the method leading to the formula rather than remembering the formula by heart!

## Line

A straight line has the following properties:

• has infinite length and the width of a point

• has a fixed position in space, unlike vectors

• is written in parametric form in all dimensions and can only be unambiguously written as an equation in two dimensions

### The parametric form of the line The parametric form for the line , or , is written as follows:

where:

• is a known point on the line that determines the position of the line

• is the direction vector of the line that determines the direction of the line

• is a parameter and assumes any value between and

The parametric form works because the expression is the sum of the local vector and the vector , whose result is exactly the point . The image shows that regardless of the value of , the result is a point on the line . Note that we need to know two points on the line to create the direction vector .

### Line equation in two dimensions The standard equation of the line is a special case because it can only be expressed in :

To construct the equation, we assume that the dot product of two perpendicular vectors is 0, in this case the normal vector and the direction vector :

Where the constants , , and are:

#### Derivation

We now derive how the constants , and are determined for the line ;

• is the normal vector to

• is a given point on

• is an arbitrary point on

The derivation is as follows:

## Plane

A plane has the following characteristics:

• is a flat, infinite and two-dimensional surface

• has a fixed position in the space

• can both be expressed in parametric form and written as an equation

### The parametric form of a plane The parametric form of a plane is:

where:

• is a given point on the plane

• and are two different vectors lying on the plane

• are parameters and assume any values between to

Three known points on the plane are enough to express the parametric form. Let , and be three known points. We can then construct the vectors as follows:

### The equation of a plane The standard equation for a plane in is:

where the constants are:

The explanation behind the constants' definitions is:

• is the normal vector to

• is a known point on

• is an arbitrary point on

The basis for the construction of the plane's equation is that all vectors on the plane have a right angle to the plane's normal vector , making the dot product between them 0. We start with and derive from it:

The generalized form of the plane's equation in is therefore:

Remember that a plane is always two-dimensional, no matter what space it is in. It is a common misconception among beginners that the plane's equation must have exactly two variables, e.g. and . But as you can see above, the plane can have several variables, up to . The plane's dimensions are still two, so it extends into space along two directions (remember the plane's parametric form). To sort out the mixture of the number of variables and dimensions, we have the following three rules to memorize:

1. A point has 0 directions and 0 dimensions

2. A line has 1 direction and 1 dimension

3. A plane has 2 directions and 2 dimensions

A practical example is to imagine an A4 paper inside a room. The room has three dimensions (width, height and depth) while the A4 paper has two dimensions (width and depth). Quite frankly, this example is deficient, partly because the paper actually has a height of a few millimeters and partly because it is neither infinite nor completely flat (it has microscopic irregularities). Fortunately, an engineer knows the noble art of effectively simplifying reality.

## Intersection

### Algebra and geometric interpretation

Examples of an intersection and its geometric interpretation can be the point where two lines intersect or the line where two planes intersect. Purely algebraically, we deal with intersections when an equation system has a solution, i.e. what point or points the equations in the system have in common. For each system of equations, one, and only one, of the following three cases always applies:

• a unique solution

• infinitely many solutions

• no solution

Here we visualize three examples - one for each of the three cases of solutions of a system of equations in : unique solution (one point), infinitely many solutions (one line) and no solution.   ### Intersection of lines in

To find the intersection of two or more lines in when the equations are given, it is easiest to set up a system of equations.

The solution , if any, is the desired intersection. We find the solution with the help of Gaussian-Jordan elimination.

### Intersection of lines in To find the intersection of two lines, and , in , it is easiest to set their parameter representations equal to each other:

and lock one of the parameters, or , by setting either equal to , for example . Then the following remains:

Now there is one equation and one variable, namely the parameter . If there is a solution, you obtain the intersection point from the parametric form for when . If you have more lines, repeat the procedure for all lines.

### Intersection between planes

To find the intersection between planes, we show here an example for the three planes , and in . We set up a system of equations:

Using Gaussian-Jordan elimination, we solve the system of equations. One, and only one, of the following three cases will then be true:

1. unique solution - intersection is a point

2. infinitely many solutions - the intersection is a line

3. no solutions - the planes do not intersect

### Intersection between a line and a plane To find the intersection between the line and the plane in , we put the parametric form of the line into the equation of the plane. Let:

Each point on the line can be expressed using its parametric form as:

and since we are searching for a common point, we can assume that such a point exists and insert the expression into the equation for :

Given that is defined, i.e. the denominator above is non-zero, the value is entered in the parametric form for to get the intersection point.

## Parametric form

Parameterization is the process of creating a parametric form that implicitly describes a line, curve or surface (also called manifold). The idea is that instead of explicitly expressing an equation with variables, one uses a parameter that can assume any value between and . Compare these two expressions:

The left expression is the parametric form that describes the same line as the equation (the expression on the right). Each value for thus implicitly expresses a unique point on the line, while the equation explicitly describes the balance between each and coordinate.  